Question

How do you find the center and radius for ${x^2} + {y^2} = 36$ ?

Answer 1

Compare the given equation to the standard equation of a circle and identify the center and the radius.
The standard equation of a circle is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ where point $\left( {h,k} \right)$ is the center and $r$ is the radius of the circle such that $r > 0$.

Complete step-by-step solution:
The given equation of the circle is ${x^2} + {y^2} = 36$.
We can convert the given equation into the standard form of a circle as shown below.
$\Rightarrow {\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {\left( 6 \right)^2}$
Now, compare the obtained equation with standard equation of a circle ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ where point $\left( {h,k} \right)$ is the center and $r$ is the radius of the circle such that $r > 0$.
It is observed that the value for $h$ and $k$ is $0$ and the value for $r$ is $6$ as $r < 0$ is not allowed.
Therefore, center of the circle is $\left( {h,k} \right) = \left( {0,0} \right)$ and radius of the circle is $r = 6$ for the given circle equation ${x^2} + {y^2} = 36$.
Circle is a close figure, uniquely defined by the position of a fixed point (center) and the constant distance between the fixed point and the point on the circle (radius).
All the possible circles in a plane are similar.

Note: Always convert the given general equation to standard equation then compare to obtain the center and radius of the circle. If we assume the position of a center is point $\left( {h,k} \right)$, any point on circle is $\left( {x,y} \right)$ and the constant distance between center $\left( {h,k} \right)$ and any point on circle $\left( {x,y} \right)$ is radius $r$ then according to distance formula $\sqrt {{{\left( {x - h} \right)}^2} + {{\left( {y - k} \right)}^2}} = r$ which is equivalent to the equation of a circle in standard form ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$.